Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 at 3:00
jefe_16
113
add a comment |
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 at 3:00
jefe_16
113
add a comment |
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
asked Nov 29 at 3:00
jefe_16
113
Full Question: Let $D_r$ be the dilation operator $D_rf(x) = f(rx)$ on $L^p(mathbb{R}^d),, 1 leq p < infty$. Show that $lim_{rto s} |D_rf-D_sf|_p = 0,,r,,s > 0$
I was told to use $int f(rx)dlambda^d(x) = |r|^{-d}int f(x)dlambda^d(x)$.
So I was thinking that set $s=1$, and prove $lim_{rto 1}|D_rf-D_1f|_p = lim_{rto 1}|f(rx)- f(x)| = 0$, then show this scales for all $s$. To show this I was thinking that since we know $|f-g|_p < epsilon$, where $g$ vanishes to $0$ outside a bound, then do something along the lines of $|D_rf - f|_p leq |D_rf-D_rg|_p+|D_rg-g|_p+|g-f|_p$, and $|D_rf-D_rg| = int(f(rx)-g(rx)^pdlambda^{1/p} to |r|^{-d/p}|f-g|_p < epsilon$, but I've been stuck here.
measure-theory lp-spaces
measure-theory lp-spaces
asked Nov 29 at 3:00
jefe_16
113
asked Nov 29 at 3:00
jefe_16
113
asked Nov 29 at 3:00
jefe_16
113
asked Nov 29 at 3:00
jefe_16
113
asked Nov 29 at 3:00
jefe_16
113
113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018094%2fshow-that-lim-r-to-s-d-rf-d-sf-p-0-r-s-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
add a comment |
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
add a comment |
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
Given $epsilon >0$ there exists $g in C_c(mathbb R^{d})$ such that $|f-g|<epsilon$. DCT tells you that the result is true with $g$ in place of $f$. Now $|f(rx)-f(x)| leq |g(rx)-g(x)|+|f(x)-g(x)|+|f(rx)-g(rx)|$ and $|f(rx)-g(rx)|=r^{-d} |f(x)-g(x)|$.
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
answered Nov 29 at 6:09
Kavi Rama Murthy
48.9k31854
48.9k31854
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018094%2fshow-that-lim-r-to-s-d-rf-d-sf-p-0-r-s-0%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
